Trial and Error – Physicspupil

November 8, 2023December 10, 2023

202311081445 Pastime Exercise 006

The figure below is a representation of the $\mathbf{E}$ -field by electric field lines in a family $R$ of infinitely many quadratic functions

$y_t(x_{t'})=a_tx_{t'}^2+b_tx_{t'}+c_t$

Roughwork.

A tangent to any quadratic curve at some point in the locus gives the $\mathrm{\pm ve}$ direction of the electric force experienced by a (positive) test charge placed there. I.e.,

$\displaystyle{T(x_{t'},y_{t})=\frac{\mathrm{d}}{\mathrm{d}x}\big( y_t(x_{t'})\big) = 2a_tx_{t'}+b_{t}}$ ,

the slope of tangent.

WLOG we work with the first quadrant. First, begin with the repulsive force $\mathrm{}_Q\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $+Q$ .

$\begin{aligned} \mathrm{}_QF_{q,x}^2+\mathrm{}_QF_{q,y}^2 & = \mathrm{}_QF_{q}^2 \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg)\mathrm{}_QF_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_QF_{q,x} & = \frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}} \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg) \Bigg(\frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_Q\mathbf{F}_{q} & = \mathrm{}_QF_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_QF_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Next, continue with the attractive force $\mathrm{}_{-Q}\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $-Q$ .

$\begin{aligned} \mathrm{}_{-Q}F_{q,x}^2+\mathrm{}_{-Q}F_{q,y}^2 & = \mathrm{}_{-Q}F_{q}^2 \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg)\mathrm{}_{-Q}F_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}F_{q,x} & = \frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}} \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg) \Bigg(\frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}\mathbf{F}_{q} & = \mathrm{}_{-Q}F_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_{-Q}F_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Adding $\mathrm{}_{Q}\mathbf{F}_{q}$ and $\mathrm{}_{-Q}\mathbf{F}_{q}$ will give the resultant electric force $\mathbf{F}_{E}(x_{t'},y_t)$ .

Note that

$\begin{aligned} \mathrm{}_QF_q & = k\frac{Q}{\mathrm{}_Qr_{q}^2} \\ \mathrm{}_{-Q}F_q & = k\frac{-Q}{\mathrm{}_{-Q}r_{q}^2} \\ \end{aligned}$

where

$\begin{aligned} \mathrm{}_Qr_q & = \sqrt{(d/2+x_{t'})^2+y_t^2} \\ \mathrm{}_{-Q}r_q & = \sqrt{(d/2-x_{t'})^2+y_t^2} \\ & \\ & \\ \end{aligned}$

are the distances of a test charge at $q(x_{t'},y_t)$ from point charges $+Q$ at $A(-d/2,0)$ and $-Q$ at $B(d/2,0)$ .

The smallest angle between $\mathbf{F}_{E}(x_{t'},y_{t})$ and the level should be equal to the slope of tangent $2a_tx_{t'}+b_{t}$ at point $q(x_{t'},y_t)$ .

As of the vertex of each parabola, the $x$ -coordinate is

$\displaystyle{0=x_{t'}=-\frac{b_{t}}{2a_{t}}}$

s.t. $b_{t}=0$ , and the $y$ -coordinate $c_{t}=y_t(0)$ .

I guess, under correction, the separation distance $d$ is none any parameter of the loci.

Visualisation is $\textrm{\scriptsize{NOT}}$ mathematical $\textrm{\scriptsize{BUT}}$ conceptual.

May 25, 2021June 24, 2023

202105251532 Homework 2 (Q1)

Prove that the electric field is always perpendicular to equipotential surface.

Solution.

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 2 Solution.)

$\mathbf{E}$ is the electric field vector; $\mathbf{dr}$ is a line element vector on the equipotential surface.

For any two arbitrary points $a$ and $b$ on the equipotential surface, we have the same potential there (i.e., $V(a)=V(b)$ ). From $-\int_a^b\mathbf{E}\cdot\mathbf{dr}=V(b)-V(a)=0$ . Thus $\mathbf{E}\cdot \mathbf{dr}=0$ , or, $\mathbf{E}\perp\mathbf{dr}$ .

May 24, 2021October 24, 2022

202105241204 Homework 1 (Q3)

Prove

(a) $\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A})-A\times (\nabla f)$

(b) $\nabla \times (\mathbf{A}\times \mathbf{B})=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Attempts. (brute force)

(a)

$\begin{aligned} & \quad \nabla\times (f\mathbf{A}) \\ & = \nabla \times (fA_x,fA_y,fA_z) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ fA_x & fA_y & fA_z \end{vmatrix} \\ & = \bigg( \frac{\partial}{\partial y}(fA_z)-\frac{\partial}{\partial z}(fA_y),\, \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z),\, \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x) \bigg) \\ & = \Bigg( \bigg( f\frac{\partial A_z}{\partial y} + \frac{\partial f}{\partial y}A_z - f\frac{\partial A_y}{\partial z} - \frac{\partial f}{\partial z}A_y \bigg) , \\ & \quad \qquad \bigg( f\frac{\partial A_x}{\partial z} + \frac{\partial f}{\partial z}A_x - f\frac{\partial A_z}{\partial x} - \frac{\partial f}{\partial x}A_z \bigg) , \\ & \qquad \qquad \bigg( f\frac{\partial A_y}{\partial x}-\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}A_x - f\frac{\partial A_x}{\partial y} \bigg) \Bigg) \\ & = f\Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg) ,\, \bigg( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial z} \bigg),\, \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\bigg) \Bigg) \\ & \quad \qquad + \bigg( \frac{\partial f}{\partial y}A_z - \frac{\partial f}{\partial z}A_y,\, \frac{\partial f}{\partial z}A_x - \frac{\partial f}{\partial x}A_z,\, \frac{\partial f}{\partial x}A_y - \frac{\partial f}{\partial y}A_x \bigg) \\ & = f(\nabla \times \mathbf{A}) + \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = f(\nabla \times \mathbf{A}) + (\nabla f)\times\mathbf{A} \\ & = f(\nabla \times \mathbf{A}) - \mathbf{A}\times (\nabla f) \\ \end{aligned}$

(b)

$\begin{aligned} \textrm{LHS}\enspace & = \nabla \times (\mathbf{A}\times \mathbf{B}) \\ & = \nabla \times (A_yB_z-A_zB_y,\, -A_xB_z+A_zB_x,\, A_xB_y-A_yB_x) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_yB_z-A_zB_y & -A_xB_z + A_zB_x & A_xB_y - A_yB_x \end{vmatrix} \\ & = \Bigg( \bigg( \frac{\partial}{\partial y}(A_xB_y) - \frac{\partial}{\partial y}(A_yB_x) - \frac{\partial}{\partial z}(A_xB_z) + \frac{\partial}{\partial z}(A_zB_x) \bigg) ,\, \\ & \quad \qquad \bigg( -\frac{\partial}{\partial x}(A_xB_y) + \frac{\partial}{\partial x}(A_yB_x) + \frac{\partial}{\partial z}(A_yB_z) - \frac{\partial}{\partial z}(A_zB_y) \bigg) ,\, \\ & \qquad \qquad \bigg( \frac{\partial}{\partial x}(-A_xB_z) - \frac{\partial}{\partial x}(A_zB_x) - \frac{\partial}{\partial y}(A_yB_z) + \frac{\partial}{\partial y}(A_zB_y) \bigg) \Bigg) \\ & = \Bigg( \bigg( A_x\frac{\partial B_y}{\partial y} + \frac{\partial A_x}{\partial y}B_y - A_y\frac{\partial B_x}{\partial y} - \frac{\partial A_y}{\partial y}B_x - A_x\frac{\partial B_z}{\partial z} - \frac{\partial A_x}{\partial z}B_z + A_z\frac{\partial B_x}{\partial z} + \frac{\partial A_z}{\partial z}B_x \bigg) ,\, \\ & \quad \qquad \bigg( -A_x\frac{\partial B_y}{\partial x} - \frac{\partial A_x}{\partial x}B_y + A_y\frac{\partial B_x}{\partial x}+\frac{\partial A_y}{\partial x}B_x + A_y\frac{\partial B_z}{\partial z} + \frac{\partial A_y}{\partial z}B_z - A_z\frac{\partial B_y}{\partial z} - \frac{\partial A_z}{\partial z}B_y \bigg) ,\, \\ & \qquad \qquad \bigg( A_x\frac{\partial B_z}{\partial x} + \frac{\partial A_x}{\partial x}B_z - A_z\frac{\partial B_x}{\partial x} - \frac{\partial A_z}{\partial x}B_x - A_y\frac{\partial B_z}{\partial y} - \frac{\partial A_y}{\partial y}B_z + \frac{\partial A_z}{\partial y}B_y + A_z\frac{\partial B_y}{\partial y} \bigg) \Bigg) \\ \end{aligned}$

$\textrm{RHS}=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Inspect these four terms on the right hand side by expanding one after the other.

The first term being

$\begin{aligned} (\mathbf{B}\cdot\nabla )\mathbf{A} & = \bigg( B_x\frac{\partial}{\partial x} + B_y\frac{\partial}{\partial y} + B_z\frac{\partial}{\partial z} \bigg) \mathbf{A} \\ & = \bigg( B_x\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_x}{\partial y} + B_z\frac{\partial A_x}{\partial z},\, \\ & \quad \qquad B_x\frac{\partial A_y}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_y}{\partial z},\, \\ & \qquad \qquad B_x\frac{\partial A_z}{\partial x} + B_y\frac{\partial A_z}{\partial y} + B_z\frac{\partial A_z}{\partial z} \bigg)\end{aligned}$

the second term being

$\begin{aligned} (\nabla \cdot \mathbf{B})\mathbf{A} & = \bigg( \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\bigg)\mathbf{A} \\ & = \bigg( \frac{\partial B_x}{\partial x}A_x + \frac{\partial B_y}{\partial y}A_x + \frac{\partial B_z}{\partial z}A_x ,\, \\ & \quad \qquad \frac{\partial B_x}{\partial x}A_y + \frac{\partial B_y}{\partial y}A_y + \frac{\partial B_z}{\partial z}A_y , \, \\ & \qquad \qquad \frac{\partial B_x}{\partial x}A_z + \frac{\partial B_y}{\partial y}A_z + \frac{\partial B_z}{\partial z}A_z \bigg) \\ \end{aligned}$

the third term being

$\begin{aligned} -(\mathbf{A}\cdot\nabla )\mathbf{B} & = - \bigg( A_x\frac{\partial}{\partial x} + A_y\frac{\partial}{\partial y} + A_z\frac{\partial}{\partial z} \bigg) \mathbf{B} \\ & = -\bigg( A_x\frac{\partial B_x}{\partial x} + A_y\frac{\partial B_x}{\partial y} + A_z\frac{\partial B_x}{\partial z},\, \\ & \quad\qquad A_x\frac{\partial B_y}{\partial x} + A_y\frac{\partial B_y}{\partial y} + A_z\frac{\partial B_y}{\partial z},\, \\ & \qquad\qquad A_x\frac{\partial B_z}{\partial x} + A_y\frac{\partial B_z}{\partial y} + A_z\frac{\partial B_z}{\partial z} \bigg) \\\end{aligned}$

and the fourth and last term being

$\begin{aligned} -(\nabla \cdot \mathbf{A})\mathbf{B} & = -\bigg( \frac{\partial A_x}{\partial x} +\frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \bigg)\mathbf{B} \\ & = - \bigg( B_x\frac{\partial A_x}{\partial x} + B_x\frac{\partial A_y}{\partial y} + B_x\frac{\partial A_z}{\partial z} ,\, \\ & \quad\qquad B_y\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_y\frac{\partial A_z}{\partial z} ,\, \\ & \qquad \qquad B_z\frac{\partial A_x}{\partial x} + B_z\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_z}{\partial z}\bigg) \\ \end{aligned}$

One can check that $\textrm{LHS}=\textrm{RHS}$ .

Solution. (proof)

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Using Einstein summation (/notation) and the Levi-Civita symbol $\varepsilon_{ijk}$ ,

(a)

$\begin{aligned} & \quad \nabla \times (f\mathbf{A}) \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(f\mathbf{A}_k)\cdot\varepsilon_{ijk}\qquad\qquad\qquad i,j,k\in\{ x,y,z\} \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\bigg(\frac{\partial}{\partial j}f\bigg)\cdot A_k\cdot\varepsilon_{ijk}+\sum_{i,j,k}\hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_k \bigg)\cdot f\cdot \varepsilon_{ijk} \\ & = (\nabla f)\times \mathbf{A} + f\cdot (\nabla\times\mathbf{A}) \\ & = (\nabla f)\times \mathbf{A} - A\times (\nabla f) \end{aligned}$

(b)

$\begin{aligned} \mathbf{A}\times\mathbf{B} & = \sum_{k,l,m}\hat{\mathbf{e}}_kA_lB_m\varepsilon_{klm}\\ \nabla\times (\mathbf{A}\times\mathbf{B}) & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(\sum_{l,m}A_lB_m\varepsilon_{klm})\varepsilon_{ijk} \\ & = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] \varepsilon_{klm}\varepsilon_{ijk} \\ \textrm{by } & \varepsilon_{klm}\varepsilon_{ijk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \\ \textrm{Thus, }& = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \\ & = \sum_{i,j,k,l,m}\bigg[ \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m (-\delta_{im}\delta_{jl}) \\ & \quad\qquad + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}B_m\bigg) A_l\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i \bigg( \frac{\partial}{\partial j}B_m \bigg) A_l (-\delta_{im}\delta_{jl}) \bigg] \\ & = (\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B} \end{aligned}$

QED

and the proof is more concise.

May 24, 2021January 19, 2023

202105241144 Homework 1 (Q4)

$\displaystyle{\int \delta (f(x))g(x)\,\mathrm{d}x}=\, ?$

Attempts.

$\begin{aligned} &\quad \int \delta(f(x))g(x)\,\mathrm{d}x \\ & = \int\delta (f(x))g(f^{-1}f(x))\frac{\mathrm{d}f(x)}{f'(x)} \\ & = \frac{g(f^{-1}f(x))}{f'(x)}\bigg|_{f(x)=0} \\ & = \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \end{aligned}$

Solution.

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Let the roots to $f(x)=0$ be $a_i$ where $i=1,2,3,\dots$ .
From $y=f(x)$ (or, $x=f^{-1}(y)$ ), we have $\mathrm{d}y=f'(x)\,\mathrm{d}x$ .

$\begin{aligned} &\quad \int \delta (f(x))g(x)\,\mathrm{d}x \\ & = \sum_i \int\delta (f(x))g(x)\,\mathrm{d}x\\ & = \sum_i \int\frac{\delta (y)g(f^{-1}(y))}{f'(f^{-1}(y))}\,\mathrm{d}y \\ & = \sum_i \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \\ & = \sum_i \frac{g(a_i)}{f'(a_i)} \\ \end{aligned}$

QED

May 24, 2021October 24, 2022

202105241107 Homework 1 (Q2)

Prove that the divergence of a curl is always zero; and the curl of a gradient is always zero.

Solution.

First,

Let $\mathbf{A}=A_x\,\hat{\mathbf{i}} + A_y\,\hat{\mathbf{j}} + A_z\,\hat{\mathbf{k}}$ .

$\begin{aligned} & \quad \nabla \cdot (\nabla\times \mathbf{A}) \\ & = \nabla \cdot \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = \nabla \cdot \Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg)\,\hat{\mathbf{i}} - \bigg( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \bigg)\,\hat{\mathbf{j}} + \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \bigg)\,\hat{\mathbf{k}} \Bigg) \\ & = \frac{\partial^2 A_z}{\partial x\partial y} - \frac{\partial^2 A_y}{\partial x\partial z}-\frac{\partial^2 A_z}{\partial y\partial x} + \frac{\partial^2A_x}{\partial y\partial z} + \frac{\partial^2A_y}{\partial z\partial x}-\frac{\partial^2A_x}{\partial z\partial y} \\ & = 0 \end{aligned}$

$\therefore$ The divergence of a curl is always zero.

Second,

$\nabla f(x,y,z) = \displaystyle{\bigg( \frac{\partial f(x,y,z)}{\partial x} ,\enspace \frac{\partial f(x,y,z)}{\partial y},\enspace \frac{\partial f(x,y,z)}{\partial z}\bigg)}$
$\mathbf{F}\stackrel{\textrm{def}}{=} \nabla f$

$\begin{aligned} \nabla \times \mathbf{F} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \\ & =\bigg( \frac{\partial^2f}{\partial y\partial z} - \frac{\partial^2f}{\partial z\partial y},\enspace \frac{\partial^2f}{\partial z\partial x} - \frac{\partial^2f}{\partial x\partial z},\enspace \frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x} \bigg) \\ & = 0 \\ \end{aligned}$

$\therefore$ The curl of a gradient is always zero.

May 24, 2021October 24, 2022

202105241031 Homework 1 (Q1)

Let $\boldsymbol{\lambda}\lambda$ be the separation vector from a fixed point $(x',y',z')$ to the point $(x,y,z)$ , and let $\lambda$ be its length. Show that

(a) $\nabla (\lambda^2)=2\boldsymbol{\lambda}$

(b) $\nabla (1/\lambda )=-\hat{\boldsymbol{\lambda}}/\lambda^2$

(c) What is the general formula for $\nabla (\lambda^n)$ .

Solution.

(a)

$\begin{aligned} \boldsymbol{\lambda}\lambda & = (x-x',y-y',z-z') \\ \lambda & = \textrm{length } = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \\ \lambda^2 & = (x-x')^2 + (y-y')^2 + (z-z')^2 \\ \nabla (\lambda^2)& =\Big( 2(x-x'), 2(y-y'), 2(z-z')\Big) \\ & = 2\boldsymbol{\lambda} \\ \end{aligned}$

(b)

$\begin{aligned} & \quad \nabla \bigg( \frac{1}{\lambda} \bigg) \\ & = \nabla \bigg( \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{1}{2}} \bigg) \\ & = -\frac{1}{2} \Big( (x-x')^2 + (y-y')^2 + (z-z')^2 \Big)^{-\frac{3}{2}} \cdot \nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} -\frac{1}{2}(\lambda^2)^{-\frac{3}{2}}\cdot 2\boldsymbol{\lambda} \\ & = -\lambda^{-3}\cdot \lambda\hat{\boldsymbol{\lambda}} \\ & = -\frac{\hat{\boldsymbol{\lambda}}}{\lambda^2} \\ \end{aligned}$

(c)

$\begin{aligned} \nabla (\lambda^n) & = \nabla (\lambda^2)^{\frac{n}{2}} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n}{2}-1}\cdot\nabla (\lambda^2) \\ & \stackrel{\textrm{(a)}}{=} \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\boldsymbol{\lambda} \\ & = \frac{n}{2}(\lambda^2)^{\frac{n-2}{2}}\cdot 2\lambda\hat{\boldsymbol{\lambda}} \\ & = n\lambda^{n-1}\hat{\boldsymbol{\lambda}} \\ \end{aligned}$

April 3, 2021May 17, 2023

202104031305 Homework 1 (Q1)

Given two vectors $\mathbf{A}=-4\,\hat{\mathbf{x}}-2\,\hat{\mathbf{y}}+\hat{\mathbf{z}}$ and $\mathbf{B}=5\,\hat{\mathbf{x}}-3\,\hat{\mathbf{y}}-2\,\hat{\mathbf{z}}$ .

(a) Determine the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$ .

(b) Determine the angle that the vector $\mathbf{A}$ makes with the $y$ -axis, and the angle that the vector $\mathbf{B}$ makes on the $yz$ -plane when it is projected on it.

(c) The vector $\mathbf{C}$ is the projection of vector $\mathbf{A}$ on the $xz$ -plane. Determine the unit vector along the direction of $\mathbf{C}\times \mathbf{A}$ .

(d) Express the vector $\mathbf{A}$ in terms of cylindrical coordinates, i.e., $\mathbf{A}=a_r\,\hat{\mathbf{r}}+a_\theta\,\hat{\boldsymbol{\theta}}+a_z\,\hat{\mathbf{z}}$ , with vectors $\hat{\mathbf{r}}$ , $\hat{\boldsymbol{\theta}}$ , and $\hat{\mathbf{z}}$ defined by the vector $\mathbf{A}$ . Determine the unit vectors $\hat{\mathbf{r}}$ , $\hat{\boldsymbol{\theta}}$ , and $\hat{\mathbf{z}}$ in terms of $\hat{\mathbf{x}}$ , $\hat{\mathbf{y}}$ , and $\hat{\mathbf{z}}$ , and the values of the coefficients $a_r$ , $a_\theta$ , and $a_z$ .

Solution.

(a) By the identity $\mathbf{A}\cdot\mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos\theta$ where $\theta$ is the angle between vectors $\mathbf{A}$ and $\mathbf{B}$ ,

$\begin{aligned} \mathbf{A}\cdot\mathbf{B} & = (-4)(5) + (-2)(-3) + (1)(-2) = -16 \\ |\mathbf{A}| & = \sqrt{(-4)^2 + (-2)^2 + (1)^2 } = \sqrt{21} \\ |\mathbf{B}| & = \sqrt{(5)^2 + (-3)^2 + (-2)^2 } =\sqrt{38} \end{aligned}$

$\begin{aligned} \cos\theta & = \frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|} \\ & = \frac{(-16)}{(\sqrt{21})(\sqrt{38})} \\ & = \frac{-16}{\sqrt{798}} \\ \theta & = \arccos \bigg( \frac{-16}{\sqrt{798}} \bigg) \\ & \Bigg[ \textrm{\scriptsize{OR}} \quad \cos^{-1}\bigg( \frac{-16}{\sqrt{798}} \bigg) \Bigg] \end{aligned}$

(b) Let the $y$ -axis be denoted by the vector $\mathbf{y}=(0,y,0)$ .

Then

$\begin{aligned} \mathbf{A}\cdot\mathbf{y} & = (-4)(0) + (-2)(y) + (1)(0) = -2y \\ |\mathbf{A}| & = \sqrt{(-4)^2+(-2)^2+(1)^2} = \sqrt{21}\\ |\mathbf{y}| & = \sqrt{(0)^2+(y)^2+(0)^2} = |y|\\ \end{aligned}$

The angle $\theta$ between vector $\mathbf{A}$ and $y$ -axis $\mathbf{y}$ is given by

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{A}\cdot\mathbf{y}}{|\mathbf{A}||\mathbf{y}|} \bigg) \\ & = \cos^{-1}\bigg(\frac{-2y}{(\sqrt{21})(|y|)}\bigg) \\ & = \cos^{-1}\bigg(\frac{-2}{\sqrt{21}}\cdot \textrm{sgn}(y)\bigg) \\ \end{aligned}$

where the sign function $\textrm{sgn}(y)$ is defined below

$\textrm{sgn}(y) = \begin{cases} +1 & \textrm{for\enspace}y\geqslant 0 \\ -1 & \textrm{for\enspace}y<0 \end{cases}$

Let the $yz$ -plane be denoted by the plane of vectors $\mathbf{p}=(0,y,z)$ for any $y,z\in\mathbb{R}$ .

Then,

$\begin{aligned} \mathbf{B}\cdot\mathbf{y} & = (5,-3,-2)\cdot (0,y,z) \\ & = (5)(0)+(-3)(y)+(-2)(z) \\ & = -3y-2z \\ |\mathbf{B}| & = \sqrt{38} \\ |\mathbf{p}| & = \sqrt{y^2+z^2} \\ \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{p}}{|\mathbf{B}||\mathbf{p}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{-3y-2z}{\sqrt{(38)(y^2+z^2)}} \bigg) \end{aligned}$

Hold on, I smell a rat. Let’s go another way round. Find the angle $\phi$ between vector $\mathbf{B}$ and the $x$ -axis $\mathbf{x}=(x,0,0)$ .

As a matter of routine, take dot product on $\mathbf{B}$ and $\mathbf{x}$ .

$\begin{aligned} \mathbf{B}\cdot\mathbf{x} & = (5,-3,-2)\cdot (x,0,0) \\ & = (5)(x)+(-3)(0)+(-2)(0) \\ & = 5x \\ |\mathbf{B}| & = \sqrt{38} \\ |\mathbf{x}| & = \sqrt{x^2} = |x| \\ \phi & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{x}}{|\mathbf{B}||\mathbf{x}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{5x}{(\sqrt{38})(|x|)}\bigg) \\ & = \cos^{-1}\bigg( \frac{5}{\sqrt{38}}\cdot\textrm{sgn}(x) \bigg) \end{aligned}$

The angle that the vector $\mathbf{B}$ makes on the $yz$ -plane is thus $\theta =90^\circ - \phi$ .

Back to the original line of thought. I.e.,

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{p}}{|\mathbf{B}||\mathbf{p}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{-3y-2z}{\sqrt{(38)(y^2+z^2)}} \bigg) \end{aligned}$

When the vector $\mathbf{B}$ is projected onto the $yz$ -plane (i.e., $\mathbf{p}$ ), the angle of projection $\theta$ is defined the angle between $\mathbf{B}$ and its orthogonal projection $\textrm{proj}_{\mathbf{p}}\mathbf{B}$ on that plane. It is clear that $\textrm{proj}_{\mathbf{p}}\mathbf{B}=(0,-3,-2)$ .

The more careful should have I written

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\textrm{proj}_{\mathbf{p}}\mathbf{B}}{|\mathbf{B}||\textrm{proj}_{\mathbf{p}}\mathbf{B}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{(5,-3,-2)\cdot (0,-3,-2)}{(\sqrt{(5)^2+(-3)^2+(-2)^2})(\sqrt{(0)^2+(-3)^2+(-2)^2})} \bigg) \\ & = \cos^{-1}\bigg( \frac{13}{(\sqrt{38})(\sqrt{13})} \bigg) \\ & = \cos^{-1}\bigg(\sqrt{\frac{13}{38}} \bigg) \end{aligned}$

As $\theta=90^\circ - \phi$ , or, $\theta +\phi =90^\circ$

$\cos (\theta +\phi )= \cos 90^\circ = 0$ .

One should check that

$\cos\theta\cos\phi - \sin\theta\sin\phi = 0$ .

By brute force

$\begin{aligned} \textrm{LHS} & = \cos\theta\cos\phi - \sin\theta\sin\phi \\ & = \bigg( \frac{\sqrt{13}}{\sqrt{38}}\bigg)\bigg( \frac{5}{\sqrt{38}}\bigg) - \bigg( \frac{\sqrt{(\sqrt{38})^2-(\sqrt{13})^2}}{\sqrt{38}}\bigg) \bigg( \frac{\sqrt{(\sqrt{38})^2-(5)^2}}{\sqrt{38}}\bigg) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}$

(c) $\mathbf{C}=(-4,0,1)$

$\begin{aligned} \mathbf{C} \times \mathbf{A} & = \begin{bmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ -4 & 0 & 1 \\ -4 & -2 & 1 \end{bmatrix} \\ & = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix} \hat{\mathbf{x}} - \begin{pmatrix} -4 & 1 \\ -4 & 1 \end{pmatrix} \hat{\mathbf{y}} + \begin{pmatrix} -4 & 0 \\ -4 & -2 \end{pmatrix} \hat{\mathbf{z}} \\ & = 2\,\hat{\mathbf{x}} + 8\,\hat{\mathbf{z}} \\ & = (2,0,8) \end{aligned}$

Thus,

$\begin{aligned} \hat{\mathbf{n}} & = \frac{\mathbf{C}\times\mathbf{A}}{|\mathbf{C}\times\mathbf{A}|} \\ & = \frac{(2,0,8)}{\sqrt{(2)^2+(0)^2+(8)^2}} \\ & = (\frac{2}{\sqrt{70}} , 0, \frac{8}{\sqrt{70}}) \end{aligned}$

(d)

$\begin{aligned} \hat{\mathbf{r}} & = \hat{\mathbf{x}}\cos\theta + \hat{\mathbf{y}}\sin\theta \\ \hat{\boldsymbol{\theta}} & = -\hat{\mathbf{x}}\sin\theta + \hat{\mathbf{y}}\cos\theta \\ \hat{\mathbf{z}} & = \hat{\mathbf{z}} \end{aligned}$

$\begin{aligned} \cos\theta & =\displaystyle{\frac{-4}{\sqrt{(-4)^2+(-2)^2}}=\frac{-4}{\sqrt{20}}} \\ \sin\theta & = \displaystyle{\frac{-2}{\sqrt{(-4)^2+(-2)^2}}=\frac{-2}{\sqrt{20}}} \end{aligned}$
$\begin{aligned} \hat{\mathbf{r}} & = -\frac{2\sqrt{5}}{5}\,\hat{\mathbf{x}} -\frac{\sqrt{5}}{5}\,\hat{\mathbf{y}}\\ \hat{\boldsymbol{\theta}} & = \frac{\sqrt{5}}{5}\,\hat{\mathbf{x}} -\frac{2\sqrt{5}}{5}\,\hat{\mathbf{y}} \\ \hat{\mathbf{z}} & = \hat{\mathbf{z}} \end{aligned}$

$\therefore \quad \mathbf{A}=2\sqrt{5}\,\hat{\mathbf{r}}+\hat{\mathbf{z}}$ .